The integrating factor of the differential eq | vedclass

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Question

(B) The given differential equation is $x \log x \frac{dy}{dx} + y = 2 \log x$. Dividing both sides by $x \log x$, we get: $\frac{dy}{dx} + \frac{1}{x

Vedclass · Accepted Answer

$\log x$

Vedclass · Answer

(B) The given differential equation is $x \log x \frac{dy}{dx} + y = 2 \log x$. Dividing both sides by $x \log x$, we get: $\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{x \log x}$. The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx}$. $IF = e^{\int \frac{1}{x \log x} dx}$. Let $u = \log x$, then $du = \frac{1}{x} dx$. $IF = e^{\int \frac{1}{u} du} = e^{\log u} = u = \log x$. Thus, the integrating factor is $\log x$.

The integrating factor of the differential equation $x \log x \frac{dy}{dx} + y = 2 \log x$ is given by

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